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# S95-05

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Math Expert
Joined: 02 Sep 2009
Posts: 51218

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16 Sep 2014, 00:49
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Difficulty:

15% (low)

Question Stats:

76% (00:41) correct 24% (00:39) wrong based on 58 sessions

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If $$p$$, $$q$$, $$x$$, and $$y$$ are positive integers, is $$\frac{q^y}{p^x}$$ an integer?

(1) $$q$$ is evenly divisible by $$p$$

(2) $$y \ge x$$

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16 Sep 2014, 00:49
1
Official Solution:

The question asks if we can determine whether the expression $$\frac{q^y}{p^x}$$ yields an integer. This is the same as asking if $$q^y$$ is divisible by $$p^x$$ or, equivalently, if $$p^x$$ is a factor of $$q^y$$.

Statement 1 tells us that $$q$$ is evenly divisible by $$p$$, or, in other words, that $$p$$ is a factor of $$q$$. Let us plug in numbers that satisfy this condition and see if we can answer the question in the prompt. We will choose $$p = 2$$ and $$q = 4$$. However, since we are given no information on $$x$$ or $$y$$, we can plug in any integer values for $$x$$ and $$y$$. If $$x = 3$$ and $$y = 1$$, then $$\frac{q^y}{p^x} = \frac{4^1}{2^3} = \frac{4}{8} = \frac{1}{2}$$, which is not an integer. On the other hand, if $$y = 3$$ and $$x = 1$$, $$\frac{q^y}{p^x} = \frac{4^3}{2^1} = \frac{64}{2} = 32$$, which is an integer. Statement 1 alone is NOT sufficient to answer the question. Eliminate answer choices A and D. The correct answer choice is B, C, or E.

Statement 2 tells us that $$y \ge x$$. However, we know nothing about the values of $$p$$ and $$q$$, which can be any integers. Whether the expression yields an integer depends on the relationship between $$p$$ and $$q$$. Statement 2 alone is also NOT sufficient to answer the question. Eliminate answer choice B. The correct answer choice is either C or E.

Taking both statements together: Statement 1 tells us that $$p$$ is a factor of $$q$$, which implies that $$p \le q$$, and statement 2 tells us that $$x \le y$$. This means that the term $$q^y$$ will always be larger than $$p^x$$, and since $$p$$ is a factor of $$q$$, $$\frac{q^y}{p^x}$$ will always yield an integer. For example, if $$p = 3$$, $$q = 6$$, $$x = 1$$, and $$y = 2$$, then $$\frac{q^y}{p^x} = \frac{6^2}{3^1} = \frac{36}{3} = 12$$. Both statements together are sufficient.

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03 Feb 2017, 09:18
if in this solution q = 6, p = 3 , y = 3 and x = x (y>=x). Then answer is 2 raised to power of 1.5, which is not integer. so answer should be neither is sufficient.
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03 Feb 2017, 11:57
kriti2016 wrote:
if in this solution q = 6, p = 3 , y = 3 and x = x (y>=x). Then answer is 2 raised to power of 1.5, which is not integer. so answer should be neither is sufficient.

Please note that the OA for this question is C, not E. So, you must be wrong somewhere. If you clarify what you mean by the red part I might be able to tell you where you went wrong.
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30 Apr 2017, 06:19
kriti2016 you would not get a fraction for the power ie. 1.5 since all variables are intergers
Re: S95-05 &nbs [#permalink] 30 Apr 2017, 06:19
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# S95-05

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